Torsten Sillke, BI, 199?
Tiling the 3-cube with equal quasi-9-cubes:
-------------------------------------------
The only possibility is the trivial 1x3x3 box.
There are some cases to check. I didn't found a short proof so far.
A packing with three congruent (reflections allowed)
quasi-9-cubes is impossible too.
As David Singmaster once said, that he had heard of another
solution which needed reflection, I have redone my analysis
by program testing all quasi-9-cubes fitting into a 2x3x3 box.
This is sufficient if you consider the center cube.
But no other quasi-9-cube packs the 3-cube.
> Return-Path:
> Date: Sat, 28 Mar 1998 20:57:27 -0500
> From: michael reid
> Reply-To: nobnet@iijnet.or.jp
> Subject: [NOBNET 791] angus lavery's challenge refuted
> Sender: owner-nobnet@iijnet.or.jp
> To: nobnet@iijnet.or.jp
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>
> david singmaster writes
>
> > A somewhat different dissection problem for the 3x3x3 cube was posed
> > by Angus Lavery some years ago, but I have never found a solution. The problem
> > is to dissect the 3x3x3 into three congruent polycube pieces (so the unit
> > cubes are not divided). Angus asserts it can be done, with one piece being a
> > reflection of the others.
>
>
> i think i can prove that the only solution is three 1x3x3 blocks,
> even allowing disconnected shapes. this is a long and torturous
> argument, and therefore somewhat susceptible to gaps and omissions.
>
> claim 1. the bounding box of such a shape cannot be 3x3x3.
> proof. suppose the bounding box is 3x3x3. then either the shape
> covers the central cell of the bounding box, or it doesn't.
> in the first case, all three pieces cover the central cell
> of the 3x3x3 cube. in the second case, none of the three
> pieces cover the central cell. both are contradictions,
> so the claim is proved.
>
> claim 2. the bounding box of such a shape is either 1x3x3 or 2x3x3.
> proof. one of the three pieces must cover at least 3 of the 8
> corners of the cube. thus it must have width 3 in at least
> two different directions. therefore the bounding box is
> either 1x3x3 or 2x3x3, as claimed.
>
> if its bounding box is 1x3x3, then the shape is a 1x3x3 box. so we
> only need to consider the case where the bounding box is 2x3x3.
> label the cells of the bounding box as shown
>
> aba ded
> bcb efe
> aba ded
>
> note that there are two different positions of the bounding box inside
> the 3x3x3 cube; they correspond to the "abc" layer being on an outer
> layer of the cube, or the "def" layer being on an outer layer.
>
> claim 3. both positions must be used.
> proof. suppose, to the contrary, that only one of these positions
> is used, say with "abc" on an outer layer of the cube.
> if the shape covers cell "f", then all three pieces cover
> the central cell of the cube, a contradiction. on the other
> hand, if the shape does not cover cell "f", then none of the
> three pieces cover the central cell, again a contradiction.
>
> so now we know that one of the positions occurs twice and the other
> occurs once. there is no loss of generality in supposing that "abc"
> on the outer layer occurs twice, and "def" on the outer layer occurs
> once. suppose that the shape has A cells of type "a" , B cells of
> type "b", and so forth. now we count the number of corner cells, edge
> cells, face cells and central cells of the 3x3x3 cube. this gives the
> equations
>
> 2 A + D = 8 (counting corners)
> A + 2 B + 2 D + E = 12 (counting edges)
> B + 2 C + 2 E + F = 6 (counting faces)
> C + 2 F = 1 (central cell)
>
> the last equation gives C = 1 and F = 0 , and then ... (hey, don't
> we all know how to do these?) ... we find that the only solutions in
> non-negative integers are
>
> A B C D E F
> case 1: 4 4 1 0 0 0
> case 2: 3 2 1 2 1 0
> case 3: 2 0 1 4 2 0
>
> we have three cases to analyze.
>
> in case 1, the shape is the 1x3x3 block.
>
> in case 2, consider the way the eight corners are covered. there are
> three possibilities (up to symmetry) where the pieces are labeled
> x, y and z.
>
> x.x ... y.y
> situation 1: ... ... ...
> x.z ... y.z
>
>
> x.y ... y.y
> situation 2: ... ... ...
> x.z ... x.z
>
>
> x.y ... z.y
> situation 3: ... ... ...
> x.y ... x.z
>
> in situation 1, piece z cannot reach the cell labeled t, so this cell
>
> x.x t.. y.y
> ... ... ...
> x.z ... y.z
>
> is filled by either piece x or piece y. without loss of generality,
> suppose it is filled by piece x. the corresponding cell of piece y
> must then also fill cell t, a contradiction.
>
> in situation 2, the bounding box of piece z is either parallel to the
> bounding box of piece x, or parallel to the bounding box of piece y.
> these two cases are equivalent by switching the roles of x and y.
> so suppose the bounding box of piece z is as shown
>
> .zz .zz .zz
> .zz .zz .zz
> .zz .zz .zz
>
> then of the cells labeled t, 3 are covered by piece x (2 cells of
> type "d" and 1 of type "e"), 6 are covered by piece z (3 of type "a",
) .t. .t. .t.
) .t. .t. .t.
) .t. .t. .t.
> 2 of type "b" and 1 of type "c") and at least 1 is covered by piece y
> (cell of type "c"). however, there are only 9 cells labeled t,
> a contradiction.
>
> in situation 3, the cell labeled t cannot be filled by piece z, since
>
> x.y .v. z.y
> ... v.v .t.
> x.y .u. x.z
>
> it has no cells of type "f". now there is no loss of generality in
> supposing it is filled by piece x. the corresponding cell of piece y
> must be in cell t or cell u; cell t is occupied, so it must be in
> cell u. of the cells labeled v, 1 is filled by piece x (cell of
> type "c"), 1 is filled by piece y (cell of type "c"), and 2 are filled
> by piece z (cells of type "b"). however, there are only 3 cells
> labeled v, a contradiction.
>
> this finishes case 2.
>
> in case 3, consider the way the eight corners are covered. there are
> two possibilities, up to symmetry
>
> z.z ... x.y
> situation 1: ... ... ...
> z.z ... y.x
>
>
> z.z ... x.y
> situation 2: ... ... ...
> z.z ... x.y
>
> in situation 1, the cell labeled t must be filled by piece x, from
>
> z.z ... x.y
> ... ... .t.
> z.z ... y.x
>
> the cell of type "c". similarly, it must also be filled by piece y,
> a contradiction.
>
> in situation 2, there are 3 possibilities (up to symmetry) for the
> bounding boxes of pieces x and y:
>
> xx. xx. xx. .yy .yy .yy
> subcase a: xx. xx. xx. .yy .yy .yy
> xx. xx. xx. .yy .yy .yy
>
>
> ... xxx xxx ... yyy yyy
> subcase b: ... xxx xxx ... yyy yyy
> ... xxx xxx ... yyy yyy
>
>
> xx. xx. xx. ... yyy yyy
> subcase c: xx. xx. xx. ... yyy yyy
> xx. xx. xx. ... yyy yyy
>
> in subcase a, of the cells labeled t, 6 are covered by piece x,
>
> .t. .t. .t.
> .t. .t. .t.
> .t. .t. .t.
>
> and 6 are also covered by piece y, a contradiction.
>
> in subcase b, of the cells labeled t, 6 are covered by piece x,
>
> ... ttt ...
> ... ttt ...
> ... ttt ...
>
> and 6 are also covered by piece y, again a contradiction.
>
> in subcase c, of the cells labeled t, at least 1 is covered by
>
> ... ttt ...
> ... ttt ...
> ... ttt ...
>
> piece x, from its cell of type "c". 6 of these cells are covered
> by piece y, and 3 are covered by piece z. there are only 9 cells
> labeled t, so this is a contradiction.
>
> this finishes case 3, so we're done.
>
> mike
> Return-Path:
> Date: Sun, 29 Mar 1998 00:05:15 -0500
> From: michael reid
> Reply-To: nobnet@iijnet.or.jp
> Subject: [NOBNET 792] angus lavery's challenge refuted
> Sender: owner-nobnet@iijnet.or.jp
> To: nobnet@iijnet.or.jp
> Errors-To: owner-nobnet@iijnet.or.jp
> Message-Id: <199803290505.OAA08457@ml0.iijnet.or.jp>
> X-Sequence: NOBNET 792
> Precedence: bulk
> Lines: 28
>
> rereading my post, i see that i omitted a diagram:
>
> (from case 2)
>
> > in situation 2, the bounding box of piece z is either parallel to the
> > bounding box of piece x, or parallel to the bounding box of piece y.
> > these two cases are equivalent by switching the roles of x and y.
> > so suppose the bounding box of piece z is as shown
> >
> > .zz .zz .zz
> > .zz .zz .zz
> > .zz .zz .zz
> >
> > then of the cells labeled t, 3 are covered by piece x (2 cells of
> > type "d" and 1 of type "e"), 6 are covered by piece z (3 of type "a",
> > 2 of type "b" and 1 of type "c") and at least 1 is covered by piece y
> > (cell of type "c"). however, there are only 9 cells labeled t,
> > a contradiction.
>
> the missing diagram should be
>
> .t. .t. .t.
> .t. .t. .t.
> .t. .t. .t.
>
> which shows where the cells labeled t are.
>
> mike